WBJEE 2019 Chemistry Question Paper with Answer and Solution

(B) The given equation of the hyperbola is $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$.
Comparing this with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.
The coordinates of the foci are given by $(\pm ae, 0)$,where $ae = \sqrt{a^2 + b^2}$.
Substituting the values,$ae = \sqrt{\cos^2 \alpha + \sin^2 \alpha} = \sqrt{1} = 1$.
Thus,the foci are $(\pm 1, 0)$,which are independent of $\alpha$.
Therefore,the foci remain constant as $\alpha$ varies.

(A) Key point: As the percentage of $s$-character in a bond increases,the bond angle also increases.
According to Bent's Rule,the lone pair occupies an orbital with more $s$-character as the electronegativity of the central atom increases.
In $NH_3$,the $N$ atom is more electronegative than $P$ in $PH_3$.
Since the bond angle in $NH_3$ $(107.6^{\circ})$ is greater than in $PH_3$ $(93.5^{\circ})$,the bonding orbitals in $NH_3$ have more $s$-character.
Consequently,the lone pair orbital in $NH_3$ has more $s$-character and less $p$-character compared to the lone pair in $PH_3$.
Thus,the $p$-character of the lone pair on ammonia is less than that of phosphine.
Hence,option $(A)$ is the correct answer.

(C) In the equilibrium $H_{2} + I_{2} \rightleftharpoons 2 HI$,if at a given temperature,the concentrations of the reactants are increased,the value of the equilibrium constant $K_{C}$ will remain the same.
This is because the equilibrium constant $K_{C}$ is a function of temperature only and does not depend on the initial molar concentrations of the reactants or products.
The correct option is $C$.

(B) Key Point:
$(i)$ Half-filled or fully-filled configurations have low electron affinity (i.e.,positive or less negative electron gain enthalpy).
$(ii)$ Smaller the size,generally the more negative the electron gain enthalpy.
On moving across a period,the size of atoms decreases due to an increase in nuclear charge.
$(i)$ Electronic configurations of the elements are:
$C (Z=6) = 1s^{2} 2s^{2} 2p^{2}$
$N (Z=7) = 1s^{2} 2s^{2} 2p^{3}$
$O (Z=8) = 1s^{2} 2s^{2} 2p^{4}$
$(ii)$ Due to the stable half-filled electronic configuration of nitrogen $(2p^{3})$,it has a very low electron affinity (positive electron gain enthalpy value of $30.9 \ kJ/mol$).
$(iii)$ As the size of $O$ is smaller than that of $C$,$O$ has a more negative electron gain enthalpy $(-141.1 \ kJ/mol)$ compared to $C$ $(-122.3 \ kJ/mol)$.
Thus,the correct order of electron affinity is $N < C < O$.

(B) Key Point: $A$ compound with more ionic character has a higher melting point.
$Be$ and $Ca$ belong to the same group $(2)$,and ionic character increases down the group due to the decrease in polarizing power of the cation.
Thus,$BeCl_2$ is more covalent (less ionic) than $CaCl_2$,so $CaCl_2$ has a higher melting point than $BeCl_2$.
$HgCl_2$ is highly covalent due to the pseudo-noble gas configuration of $Hg^{2+}$ ($18$ electrons in the outer shell),which exerts a high polarizing power on the chloride ion.
Therefore,the order of melting points is $CaCl_2 > BeCl_2 > HgCl_2$,which corresponds to $(ii) > (i) > (iii)$.
This is equivalent to $iii < i < ii$.

(B) The charge on $1 \text{ mole}$ of electrons is $F = 96500 \text{ C/mol}$.
Given,$1 \text{ millimole} = 1 \times 10^{-3} \text{ mol}$ of $M^{n+}$ ions.
The total charge $Q = n \times \text{moles} \times F$.
$193 = n \times (1 \times 10^{-3}) \times 96500$.
$193 = n \times 96.5$.
$n = \frac{193}{96.5} = 2$.
Thus,the value of $n$ is $2$.

(C) lone pair of electrons is considered delocalised if it is in conjugation with a $\pi$-bond (i.e.,separated by one single bond).
In option $(a)$,the oxygen lone pair is separated from the carbonyl group by a $-CH_2-$ group,so it is localised.
In option $(b)$,the nitrogen lone pair is separated from the carbonyl group by a $-CH_2-$ group,so it is localised.
In option $(c)$,the oxygen atom is directly attached to the double bond of the ring. The lone pair on oxygen is in conjugation with the $\pi$-electrons of the double bond,making it delocalised.
In option $(d)$,the nitrogen lone pair is separated from the double bond by a $-CH_2-$ group,so it is localised.
Therefore,only the molecule in option $(c)$ has a delocalised lone pair.

(A) nucleophile is a species that can donate an electron pair to form a chemical bond.
In $BH_{4}^{-}$,the boron $(B)$ atom has a formal negative charge but it has no lone pair of electrons available for donation.
In $CH_{3}MgBr$,the carbon $(C)$ atom is nucleophilic due to the $C-Mg$ bond polarity.
In $CH_{3}OH$,the oxygen $(O)$ atom has two lone pairs.
In $CH_{3}NH_{2}$,the nitrogen $(N)$ atom has one lone pair.
Therefore,the $B$ atom in $BH_{4}^{-}$ is not a nucleophilic site.
Hence,the correct option is $A$.

(D) The conformations of $n$-butane (eclipsed,gauche,and anti) are generated by the rotation around the $C2-C3$ sigma bond. This rotation changes the dihedral angle between the two methyl groups attached to the $C2$ and $C3$ carbons,leading to different energy states.
Hence,the correct option is $(d)$.

(D) Heating $\beta$-keto acids or substituted malonic acids leads to decarboxylation (loss of $CO_2$).
For a compound to become achiral after decarboxylation,the resulting product must have a plane of symmetry,a center of symmetry,or be a molecule where the chiral center is destroyed (i.e.,the carbon atom becomes bonded to two identical groups).
In option $(D)$,the starting material is $2$-ethyl-$3$-oxobutanoic acid derivative. Upon heating,it undergoes decarboxylation to form $2$-butanone (or a similar derivative depending on the specific structure). Specifically,the structure in $(D)$ is $2$-ethyl-$3$-oxobutanoic acid. Upon decarboxylation,the chiral center at the $\alpha$-carbon (which has $H$,$Et$,$CO_2H$,and $COCH_3$ groups) loses the $CO_2H$ group and gains an $H$ atom. If the resulting product has two identical groups attached to the central carbon (e.g.,two $H$ atoms or two $Et$ groups),it becomes achiral.
Looking at the provided solution image,the transformation in $(d)$ results in a molecule where the central carbon is bonded to $H$,$Et$,$Me$,and $H$ (if the $CO_2H$ is replaced by $H$). Wait,the image shows the product of $(d)$ as having $H$,$Et$,$Me$,and $H$ attached to the central carbon,making it achiral because it has two identical $H$ atoms.

(C) To determine the lowest $pH$,we analyze the nature of each mixture:
$(A)$ $10 \ mL$ $0.05 \ N$ $CH_{3}COOH$ $(0.5 \ mmol)$ + $5 \ mL$ $0.1 \ N$ $NH_{4}OH$ $(0.5 \ mmol)$ forms a salt of weak acid and weak base $(CH_{3}COONH_{4})$,which is nearly neutral.
$(B)$ $5 \ mL$ $0.2 \ N$ $NH_{4}Cl$ + $5 \ mL$ $0.2 \ N$ $NH_{4}OH$ forms a basic buffer $(pH > 7)$.
$(C)$ $5 \ mL$ $0.1 \ N$ $CH_{3}COOH$ $(0.5 \ mmol)$ + $10 \ mL$ $0.05 \ N$ $CH_{3}COONa$ $(0.5 \ mmol)$ forms an acidic buffer where $pH = pK_{a} + \log(\frac{[Salt]}{[Acid]})$. Since $[Salt] = [Acid]$,$pH = pK_{a} \approx 4.76$.
$(D)$ $5 \ mL$ $0.1 \ N$ $CH_{3}COOH$ + $5 \ mL$ $0.1 \ N$ $NaOH$ forms $CH_{3}COONa$,which undergoes anionic hydrolysis,resulting in a basic solution $(pH > 7)$.
Comparing these,the acidic buffer in option $(C)$ has the lowest $pH$.

(B) In halogen acids,as the size of the halogen atom increases,the bond strength between the halogen and hydrogen atom decreases.
This makes the $H-X$ bond easier to break,facilitating the electrophilic addition reaction.
The bond dissociation energy follows the order: $HCl > HBr > HI$.
Therefore,the reactivity order for the addition reaction with ethylene is $HI > HBr > HCl$.
Thus,the correct option is $(B)$.

(C) The molar mass of the crystalline solid $MSO_{4} \cdot nH_{2}O$ is $250 \ g \ mol^{-1}$.
Given that the percentage of anhydrous salt $(MSO_{4})$ is $64\%$,the percentage of water $(H_{2}O)$ is $100\% - 64\% = 36\%$.
Mass of $H_{2}O = \frac{36}{100} \times 250 = 90 \ g$.
Moles of $H_{2}O = \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.
Since the formula is $MSO_{4} \cdot nH_{2}O$,the value of $n$ corresponds to the number of moles of water,which is $5$.
Hence,option $(C)$ is the correct answer.

(A) Given,mass of gas $(W) = 7.5 \ g$.
Volume of gas at $STP$ $(V) = 5.6 \ L$.
Since,moles of gas at $STP = \frac{V(L)}{22.4 \ L} = \frac{W}{M}$,where $M$ is the molar mass of the gas.
Therefore,$M = \frac{W \times 22.4}{V}$.
$M = \frac{7.5 \times 22.4}{5.6} = 30.00 \ g \ mol^{-1}$.
Among the given options,the molar mass $(M)$ of:
$(A) \ NO = 14 + 16 = 30.00 \ g \ mol^{-1}$.
$(B) \ N_{2}O = 28 + 16 = 44.00 \ g \ mol^{-1}$.
$(C) \ CO = 12 + 16 = 28.00 \ g \ mol^{-1}$.
$(D) \ CO_{2} = 12 + 32 = 44.00 \ g \ mol^{-1}$.
Since the molar mass of $NO$ is $30 \ g \ mol^{-1}$,the given gas is $NO$.
Hence,$A$ is the correct option.

(C) The quantity $h \nu / K_{B}$ is related to the characteristic temperature of a system.
According to the kinetic theory of gases,the average kinetic energy of a gas molecule is given by $\frac{3}{2} K_{B} T$.
Equating the energy of a photon or quantum of energy $h \nu$ to the thermal energy $K_{B} T$,we get $h \nu = K_{B} T$.
Thus,$\frac{h \nu}{K_{B}} = T$.
Therefore,the quantity $\frac{h \nu}{K_{B}}$ has the dimensions of temperature.
Hence,option $C$ is correct.

(B) The dimensions of the term $\left(\frac{a b}{V^{2}}\right)$ can be analyzed as follows:
In the van der Waals' equation,the term $\left(\frac{a}{V^{2}}\right)$ has the dimensions of pressure $(P)$.
Therefore,$\left(\frac{a}{V^{2}}\right) \times b = \text{Pressure} \times \text{Volume}$.
Since $\text{Pressure} = \frac{\text{Force}}{\text{Area}}$,we have $\text{Pressure} \times \text{Volume} = \frac{\text{Force}}{\text{Area}} \times \text{Volume} = \frac{MLT^{-2}}{L^{2}} \times L^{3} = ML^{2} T^{-2}$.
This dimension $(ML^{2} T^{-2})$ corresponds to energy (joule).
Thus,the term $\left(\frac{a b}{V^{2}}\right)$ represents energy.

(A) Let the coordinates of the fixed point $A$ be $(p, q)$ and the coordinates of the other end of the diameter $B$ be $(x, y)$.
Since $AB$ is the diameter,the center of the circle is the midpoint of $AB$,which is $(\frac{x+p}{2}, \frac{y+q}{2})$.
The radius of the circle is the distance from the center to $A$,which is $\sqrt{(\frac{x+p}{2}-p)^{2} + (\frac{y+q}{2}-q)^{2}} = \sqrt{(\frac{x-p}{2})^{2} + (\frac{y-q}{2})^{2}}$.
Since the circle touches the $X$-axis,the radius is equal to the absolute value of the $y$-coordinate of the center,so $r = |\frac{y+q}{2}|$.
Squaring both sides,we get $(\frac{x-p}{2})^{2} + (\frac{y-q}{2})^{2} = (\frac{y+q}{2})^{2}$.
Multiplying by $4$,we get $(x-p)^{2} + (y-q)^{2} = (y+q)^{2}$.
$(x-p)^{2} + y^{2} - 2qy + q^{2} = y^{2} + 2qy + q^{2}$.
$(x-p)^{2} = 4qy$.

(D) Key point: Species with one or more unpaired electrons exhibit a non-zero magnetic moment.
Electronic configurations:
$Na^{+} (Z=11) = 1s^{2} 2s^{2} 2p^{6}$ (All electrons paired)
$Mg (Z=12) = 1s^{2} 2s^{2} 2p^{6} 3s^{2}$ (All electrons paired)
$F^{-} (Z=9) = 1s^{2} 2s^{2} 2p^{6}$ (All electrons paired)
$Ar^{+} (Z=18) = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}$ (One unpaired electron in $3p$ orbital)
Since $Ar^{+}$ has an unpaired electron,it possesses a non-zero magnetic moment. Therefore,$(d)$ is the correct option.

(C) For a given value of azimuthal quantum number $l$,the magnetic quantum number $m$ can have values ranging from $-l$ to $+l$ (including zero).
If $l=0$,then $m$ must be $0$.
In option $(c)$,$l=0$ but $m=-1$,which violates the rule $-l \leq m \leq l$.
Therefore,the arrangement $n=2, l=0, m=-1$ is impossible and absurd.

(A) According to Kirchhoff's equation,the variation of enthalpy of reaction with temperature is given by:
$ \Delta H_{T_2} = \Delta H_{T_1} + \int_{T_1}^{T_2} \Delta C_{p} \, dT $
If $ \Delta C_{p} = 0 $,then the integral term becomes zero.
This implies that $ \Delta H_{T_2} = \Delta H_{T_1} $,meaning the heat of formation is independent of temperature.
Therefore,the correct condition is $ \Delta C_{p} = 0 $.
Hence,option $(A)$ is the correct answer.

(A, B) Statement $(a)$ is correct: In $CrO_{5}$ (butterfly structure),the oxidation number of $Cr$ is $+6$.
Statement $(b)$ is correct: For the reaction $N_{2}O_{4(g)} \rightarrow 2NO_{2(g)}$,$\Delta n_{g} = 2 - 1 = 1$. Since $\Delta H = \Delta U + \Delta n_{g}RT$,and $\Delta n_{g} > 0$,we have $\Delta H > \Delta U$.
Statement $(c)$ is incorrect: For $0.1 \ N \ H_{2}SO_{4}$,$[H^{+}] = 0.1 \ M$,so $pH = -\log(0.1) = 1$. For $0.1 \ N \ HCl$,$[H^{+}] = 0.1 \ M$,so $pH = -\log(0.1) = 1$. Thus,$pH$ values are equal.
Statement $(d)$ is incorrect: At $25^{\circ} C$,$\frac{RT}{F} \approx 0.0257 \ V$. The value $0.0591 \ V$ corresponds to $\frac{2.303RT}{F}$.
Therefore,the correct statements are $(a)$ and $(b)$.

(D) For the given equilibrium $H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)}:$
$(I)$ At equilibrium,the Gibbs free energy change is $\Delta G = 0$.
$(II)$ Entropy change,$\Delta S$,measures the randomness of the system. Since the vapor state is more disordered than the liquid state,the entropy increases,meaning $\Delta S > 0$.
$(III)$ The conversion of liquid $H_{2}O$ to vapor $H_{2}O$ is an endothermic process,which requires the absorption of heat energy. Therefore,the enthalpy change is $\Delta H > 0$.
Thus,the correct set of conditions is $\Delta G = 0, \Delta H > 0, \text{ and } \Delta S > 0$.
Therefore,option $(D)$ is the correct answer.

(D) The reaction of cyclopentanol with $NaH$ forms a sodium alkoxide intermediate.
This alkoxide then reacts with $CS_{2}$ to form a sodium xanthate salt.
Finally,the reaction with $CH_{3}I$ (methyl iodide) results in the formation of a methyl xanthate ester.
This sequence is known as the Chugaev elimination precursor synthesis.
Therefore,the product is a xanthate.
Thus,option $(d)$ is the correct answer.

(C) The oxidation of allyl alcohol $(CH_2=CH-CH_2OH)$ with a peracid $(RCO_3H)$ is an epoxidation reaction.
This reaction yields glycidol,which has the molecular formula $C_3H_6O_2$.
The structure of glycidol is $CH_2(O)CH-CH_2OH$ (or $2,3-$epoxypropan$-1-$ol).
In this molecule,the carbon atom at the $C-2$ position is bonded to four different groups ($-H$,$-CH_2OH$,and the two carbons of the epoxide ring),making it an asymmetric (chiral) carbon atom.
Therefore,the correct structure is $CH_2(O)CH-CH_2OH$.

(C) $2,2,2-$trichloroacetophenone reacts with aqueous $KOH$ via a haloform-like cleavage reaction. The hydroxide ion attacks the carbonyl carbon,followed by the departure of the stable $CHCl_3$ (chloroform) leaving group to form the potassium salt of benzoic acid. Subsequent acidification with $H_3O^+$ yields benzoic acid as the final product $P$.
Thus,option $(c)$ is the correct answer.

(A) The reaction of $4$-hydroxybenzoic acid with acetic anhydride in the presence of conc. $H_{2}SO_{4}$ and heat is an acetylation reaction.
The phenolic $-OH$ group acts as a nucleophile and attacks the carbonyl carbon of the acetic anhydride,leading to the formation of an ester group $(-OCOCH_{3})$.
The product $Q$ is $4$-acetoxybenzoic acid,which has the molecular formula $C_{9}H_{8}O_{4}$.
Thus,option $(A)$ is the correct answer.

(B) $CH_3COCl$ (acetyl chloride) is highly reactive and hydrolyzes to form $CH_3COOH$ (acetic acid) even at $25^{\circ} C$.
Acetic acid is a stronger acid than carbonic acid $(H_2CO_3)$,so it reacts with $NaHCO_3$ to evolve $CO_2$ gas.
The reaction is:
$CH_3COCl + H_2O \longrightarrow CH_3COOH + HCl$
$CH_3COOH + NaHCO_3 \longrightarrow CH_3COONa + H_2O + CO_2 \uparrow$
Thus,option $(B)$ is the correct answer.

(D) The rate law for the reaction is given by $r = k[A]^n$.
Taking the square root on both sides,we get $\sqrt{r} = \sqrt{k} \times [A]^{n/2}$.
The given graph is a plot of $\sqrt{r_0}$ versus $[A]_0$,which is a straight line passing through the origin.
Comparing this with the equation $y = mx$,we have $y = \sqrt{r_0}$,$x = [A]_0$,and slope $m = \sqrt{k}$.
From the graph,the slope is $4.0$.
Therefore,$\sqrt{k} = 4.0$,which implies $k = (4.0)^2 = 16.0$.
Also,comparing the powers of $[A]$,we have $n/2 = 1$,which gives $n = 2$.
Thus,the order of the reaction is $n = 2$ and the rate constant is $k = 16.0 \ dm^3 \ mol^{-1} \ min^{-1}$.
Hence,option $(d)$ is the correct answer.

(B) For a first order reaction,the relationship between half-life $(t_{1/2})$ and rate constant $(k)$ is given by $t_{1/2} = \frac{0.693}{k}$.
For reaction $(1)$: $A \rightarrow B$,$k = 0.693 \ min^{-1}$. Therefore,$t_{1/2} = \frac{0.693}{0.693} = 1.0 \ min$.
For reaction $(2)$: $A \rightarrow C$,$t_{1/2} = 0.693 \ min$. Therefore,$k = \frac{0.693}{0.693} = 1.0 \ min^{-1}$.
Comparing the rate constants,for reaction $(1)$,$k = 0.693 \ min^{-1}$ and for reaction $(2)$,$k = 1.0 \ min^{-1}$.
Since the rate of a first order reaction is $Rate = k[A]$,and the initial concentration of $A$ is the same for both,the reaction with the higher rate constant is faster.
Since $1.0 > 0.693$,reaction $(2)$ is faster than reaction $(1)$,meaning reaction $(1)$ is slower than reaction $(2)$.

(B) Let the initial radioactivity be $N_0 = 100 \%$.
Given,half-life period $(t_{1/2}) = 60$ days.
Total time $(T) = 180$ days.
The number of half-lives $(n)$ is calculated as:
$n = \frac{T}{t_{1/2}} = \frac{180}{60} = 3$.
The remaining radioactivity $(N)$ after $n$ half-lives is given by the formula:
$N = N_0 \times (\frac{1}{2})^n$.
Substituting the values:
$N = 100 \% \times (\frac{1}{2})^3 = 100 \% \times \frac{1}{8} = 12.5 \%$.
Therefore,the radioactivity after $180$ days will be $12.5 \%$.

(D) The initial nucleus is ${}_{82}A^{210}$ and the final nucleus is ${}_{82}D^{206}$.
The change in mass number is $210 - 206 = 4$ units,and the change in atomic number is $82 - 82 = 0$.
Emission of one $\alpha$-particle decreases the mass number by $4$ and the atomic number by $2$.
Emission of one $\beta$-particle keeps the mass number constant and increases the atomic number by $1$.
To keep the atomic number constant while decreasing the mass number by $4$,we need one $\alpha$-particle ($-2$ in atomic number) and two $\beta$-particles ($+2$ in atomic number).
The sequence is ${}_{82}A^{210}$ $\xrightarrow{-\beta} {}_{83}B^{210}$ $\xrightarrow{-\beta} {}_{84}C^{210}$ $\xrightarrow{-\alpha} {}_{82}D^{206}$.
Thus,the sequence of emission is $\beta, \beta, \alpha$.

(D) coordination compound of $Co(III)$ that dissociates into $3$ ions in solution must have two ionizable ions outside the coordination sphere,given the oxidation state of cobalt is $+3$.
$(A)$ $[Co(NH_{3})_{6}]Cl_{3} \rightarrow [Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$ ($4$ ions)
$(B)$ $[Co(NH_{3})_{5}(SO_{4})]Cl \rightarrow [Co(NH_{3})_{5}(SO_{4})]^{+} + Cl^{-}$ ($2$ ions)
$(C)$ $[Co(NH_{3})_{5}Cl]SO_{4} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + SO_{4}^{2-}$ ($2$ ions)
$(D)$ $[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + 2Cl^{-}$ ($3$ ions)
Thus,the compound that dissociates into $3$ ions is $[Co(NH_{3})_{5}Cl]Cl_{2}$.
Hence,option $(D)$ is the correct answer.

(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons. For $n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$.
$(A)$ In $K_{4}[Mn(CN)_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. Thus,$n=1$.
$(B)$ In $[Fe(H_{2}O)_{6}]Cl_{3}$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $H_{2}O$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
$(C)$ In $K_{3}[FeF_{6}]$,$Fe$ is in $+3$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
$(D)$ In $K_{4}[MnF_{6}]$,$Mn$ is in $+2$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand,so no pairing occurs. Thus,$n=5$.
Therefore,the compounds with five unpaired electrons are $(B)$,$(C)$,and $(D)$.

(D) copper coin is electroplated with zinc $(Zn)$ and then heated at high temperature.
During heating,zinc atoms diffuse into the copper lattice to form an alloy.
The resulting alloy is brass,which has a characteristic golden colour.
This occurs because the zinc migrates through the copper to form the $\alpha$-form of brass alloy (where the percentage of $Cu > 65 \%$ and $Zn < 35 \%$).
$Zn + Cu \xrightarrow{\Delta} \text{Brass (Golden colour)}$.
Thus,the correct option is $(D)$.

(A) Key Point: For elements with similar electronic configuration of the outermost shell,atomic size determines the value of ionisation energy. Larger size leads to lower ionisation energy.
Electronic configuration for $II^{nd}$ $I.E$:
$Zn^{+}$ ion $(Z=30) = [Ar] 3d^{10} 4s^{1} - (1734 \ kJ/mol)$
$Cd^{+}$ ion $(Z=48) = [Kr] 4d^{10} 5s^{1} - (1631 \ kJ/mol)$
$Hg^{+}$ ion $(Z=80) = [Xe] 4f^{14} 5d^{10} 6s^{1} - (1809 \ kJ/mol)$
Since the size of $Cd^{+} > Zn^{+}$,it has lower $II^{nd}$ ionisation energy. Due to the lanthanoid contraction (i.e.,poor screening by $4f$ and $5d$ electrons),$Hg^{+}$ has a higher $II^{nd}$ ionisation energy.
Hence,the correct order is $Zn > Cd < Hg$,and option $(A)$ is the correct answer.

(D) Among the given options,$_{92}^{238}U$ is an isotope of uranium but cannot be used as a nuclear fuel.
$_{92}^{238}U$ is non-fissile,meaning it does not undergo fission by thermal neutrons.
The energy released when $_{92}^{238}U$ absorbs a neutron is insufficient to carry out nuclear fission.
Hence,$(d)$ is the correct option.

(D) During the electrolysis of aqueous $CuSO_4$ solution using $Cu$ electrodes,the following ions are present in the solution: $Cu^{2+}$,$H^{+}$,$SO_{4}^{2-}$,and $OH^{-}$.
At the cathode,$Cu^{2+}$ ions are reduced: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
At the anode,the $Cu$ electrode itself undergoes oxidation because it is an active electrode: $Cu_{(s)} - 2e^{-} \rightarrow Cu^{2+}_{(aq)}$.
Therefore,the correct option is $D$.

(D) Alanine is a chiral amino acid with one chiral center. $A$ racemic mixture contains both $(R)$ and $(S)$ enantiomers.
When two alanine molecules combine to form a linear dipeptide,the resulting molecule has two chiral centers (one from each alanine residue).
Each chiral center can exist in either $(R)$ or $(S)$ configuration.
The possible combinations for the two chiral centers in the dipeptide are $(R, R)$,$(R, S)$,$(S, R)$,and $(S, S)$.
Since all these combinations represent distinct stereoisomers,there are $2^2 = 4$ possible isomeric linear dipeptides.
Therefore,the total number of isomeric linear dipeptides is $4$,and the correct option is $(d)$.

(B) In the Bayer's process,the leaching of alumina is done by using $NaOH$.
$Al_{2}O_{3(s)} + 2 NaOH_{(aq)} \stackrel{150^{\circ}C, 35 \ atm}{\longrightarrow} 2 Na[Al(OH)_{4}] + 3 H_{2}O$
$2 Na[Al(OH)_{4}] + CO_{2} \rightarrow Al_{2}O_{3} \cdot x H_{2}O + 2 NaHCO_{3}$
$Al_{2}O_{3} \cdot x H_{2}O_{(s)} \xrightarrow{1470 \ K} Al_{2}O_{3(s)} + x H_{2}O_{(g)}$
So,the option $(b)$ is the correct answer.

(A, B) The haloform reaction is given by compounds containing the $CH_{3}CO-$ group or compounds that can be oxidized to this group (like $CH_{3}CH(OH)-$ derivatives).
$1$. $ICH_{2}COCH_{2}Ph$: This compound contains an $\alpha$-hydrogen at the $ICH_{2}$ position. Under basic conditions with $I_{2}$,the $ICH_{2}$ group can be further iodinated to $CI_{3}CO-$,which then undergoes cleavage to form iodoform $(CHI_{3})$.
$2$. $PhCOCH(OH)CH_{3}$: This is a secondary alcohol with a $CH_{3}CH(OH)-$ group. It is oxidized by $I_{2}/KOH$ to the corresponding ketone $PhCOCOCH_{3}$,which contains the $CH_{3}CO-$ group and thus gives the haloform reaction.
Therefore,both compounds $(a)$ and $(b)$ respond to the haloform reaction.

(B) Chlorine bleach is $NaOCl$. The hypochlorite ion in chlorine bleach dissociates to give nascent oxygen as shown below.
$OCl^{-} \longrightarrow [O] + Cl^{-}$
Therefore,the reactive species in chlorine bleach is $OCl^{-}$.
So,the option $(B)$ is correct.

(A, C) $NH_{4}NO_{3}$ reacts with conc. $NaOH$ to evolve pungent $NH_{3}$ gas.
It reacts with conc. $H_{2}SO_{4}$ to produce brown fumes of $NO_{2}$ (due to decomposition/redox).
It shows no visible reaction with water.
Similarly,$(NH_{4})_{2}CO_{3}$ reacts with conc. $NaOH$ to evolve pungent $NH_{3}$ gas.
It reacts with conc. $H_{2}SO_{4}$ to produce effervescence of $CO_{2}$ gas.
It shows no visible reaction with water.
Both $(a)$ and $(c)$ can distinguish the three beakers.